Applying the IVT to show a value is attained

Since $f(x) = x^3 - x$ is a polynomial, it is continuous everywhere, and in particular continuous on the closed interval $[-2, 2]$. This satisfies the hypothesis needed to apply the Intermediate Value Theorem on this interval.

Next, evaluate the endpoints: $f(-2) = (-2)^3 - (-2) = -8 + 2 = -6$, and $f(2) = 2^3 - 2 = 8 - 2 = 6$. Since $-6 < 5 < 6$, the value $N = 5$ lies strictly between $f(-2)$ and $f(2)$.

By the Intermediate Value Theorem, since $f$ is continuous on $[-2,2]$ and $5$ lies between $f(-2) = -6$ and $f(2) = 6$, there must exist at least one $c$ in $(-2, 2)$ with $f(c) = 5$. This proves the desired conclusion.